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Mathematics
Problem
The true set of values of '
K
K
K
' for which
sin
−
1
(
1
1
+
sin
2
x
)
=
K
π
6
\sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) = \frac{K\pi}{6}
sin
−
1
(
1
+
s
i
n
2
x
1
)
=
6
K
π
may have a solution is :
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